| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question requiring application of constant acceleration equations to horizontal and vertical components. Students must resolve velocities, use v = u + at in both directions, and solve simultaneous equations. While it involves multiple steps and component resolution, the techniques are routine for M2 level with no novel insight required—slightly easier than average due to the guided structure and straightforward algebra. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \([25 \times (24/25) = V\cos\theta]\) | M1 | For using \(\dot{x} = V\cos\theta\) |
| \(V\cos\theta = 24\) | A1 | [2] |
| (ii) \([25 \times (7/25) = V\sin\theta - 0.3\,g]\) | M1 | For using \(\dot{y} = V\sin\theta - gt\) |
| \(V\sin\theta = 10\) | A1 | |
| \([\tan\theta = 10/24,\ V^2 = 24^2 + 10^2]\) | M1 | For using \(\tan\theta = V\sin\theta/(V\cos\theta)\) or \(V^2 = (V\cos\theta)^2 + (V\sin\theta)^2\) |
| \(\theta = 22.6°\) | A1\(\sqrt{}\) | ft dependent on both M marks |
| \(V = 26\) | A1\(\sqrt{}\) | ft dependent on both M marks [5] |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $[25 \times (24/25) = V\cos\theta]$ | M1 | For using $\dot{x} = V\cos\theta$ |
| $V\cos\theta = 24$ | A1 | **[2]** |
| **(ii)** $[25 \times (7/25) = V\sin\theta - 0.3\,g]$ | M1 | For using $\dot{y} = V\sin\theta - gt$ |
| $V\sin\theta = 10$ | A1 | |
| $[\tan\theta = 10/24,\ V^2 = 24^2 + 10^2]$ | M1 | For using $\tan\theta = V\sin\theta/(V\cos\theta)$ or $V^2 = (V\cos\theta)^2 + (V\sin\theta)^2$ |
| $\theta = 22.6°$ | A1$\sqrt{}$ | ft dependent on both M marks |
| $V = 26$ | A1$\sqrt{}$ | ft dependent on both M marks **[5]** |
---4 A particle is projected from a point $O$ with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta$ above the horizontal. After 0.3 s the particle is moving with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\tan ^ { - 1 } \left( \frac { 7 } { 24 } \right)$ above the horizontal.\\
(i) Show that $V \cos \theta = 24$.\\
(ii) Find the value of $V \sin \theta$, and hence find $V$ and $\theta$.
\hfill \mbox{\textit{CAIE M2 2009 Q4 [7]}}